pat 1077. Kuchiguse (20)

1077. Kuchiguse (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

The Japanese language is notorious for its sentence ending particles. Personal preference of such particles can be considered as a reflection of the speaker's personality. Such a preference is called "Kuchiguse" and is often exaggerated artistically in Anime and Manga. For example, the artificial sentence ending particle "nyan~" is often used as a stereotype for characters with a cat-like personality:

Itai nyan~ (It hurts, nyan~)

Ninjin wa iyada nyan~ (I hate carrots, nyan~)

Now given a few lines spoken by the same character, can you find her Kuchiguse?

Input Specification:

Each input file contains one test case. For each case, the first line is an integer N (2<=N<=100). Following are N file lines of 0~256 (inclusive) characters in length, each representing a character's spoken line. The spoken lines are case sensitive.

Output Specification:

For each test case, print in one line the kuchiguse of the character, i.e., the longest common suffix of all N lines. If there is no such suffix, write "nai".

Sample Input 1:

3
Itai nyan~
Ninjin wa iyadanyan~
uhhh nyan~

Sample Output 1:

nyan~

Sample Input 2:

3
Itai!
Ninjinnwaiyada T_T
T_T

Sample Output 2:

nai

解：蛮费事的，虽然思路很常规，因为字符串的个数不多，所以全部逆序处理了一遍，提交的时候有一组样例一路卡，还好自己找到卡的位置了，输入3 tob uob tob，应该输出ob，最后发现循环写的有点问题，改了下总算过了。

代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    int n;
lp: while(scanf("%d",&n)==1)
    {
        string *s=new string[n+1];
        int minlen=257;
        getchar();
        for(int i=0;i<n;i++)
        {
            getline(cin,s[i]);
            if(s[i].length()<minlen) minlen=s[i].length();
        }
        //重开数组
        string *t=new string[n+1];
        for(int i=0;i<n;i++)
        {
            for(int j=s[i].length()-1;j>=0;j--)
            {
                t[i].push_back(s[i][j]);
            }
        }
        string res;
        if(minlen==0)
        {
            printf("nai\n");
            goto lp;
        }
        if(minlen==1)
        {
            for(int j=1;j<n;j++)
            {
                if(t[j][0]!=t[0][0])
                {
                    printf("nai\n");
                    goto lp;
                }
            }
            printf("%c\n",t[0][0]);
            goto lp;
        }
        for(int i=1;i<minlen;i++)
        {
            string hhd=t[0].substr(0,i);
            int flag=1;
            for(int j=1;j<n;j++)
            {
                if(t[j].substr(0,i)!=hhd)
                {
                    if(i==1)
                    {
                        printf("nai\n");
                        goto lp;
                    }
                    flag=0;
                    break;
                }
            }
            if(flag==1)
            {
                res=hhd;
            }
            else
            {
                res=t[0].substr(0,i-1);
                break;
            }
        }
        for(int i=res.length()-1;i>=0;i--)
                cout<<res[i];
        printf("\n");
        goto lp;
    }
}

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